Question: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}-8x-2y &= 6 \\ -3x-2y &= -9\end{align*}$
Answer: Begin by moving the $y$ -term in the second equation to the right side of the equation. $-3x = 2y-9$ Divide both sides by $-3$ to isolate $x$ $x = {-\dfrac{2}{3}y + 3}$ Substitute this expression for $x$ in the first equation. $-8({-\dfrac{2}{3}y + 3}) - 2y = 6$ $\dfrac{16}{3}y - 24 - 2y = 6$ Simplify by combining terms, then solve for $y$ $\dfrac{10}{3}y - 24 = 6$ $\dfrac{10}{3}y = 30$ $y = 9$ Substitute $9$ for $y$ in the top equation. $-8x-2( 9) = 6$ $-8x-18 = 6$ $-8x = 24$ $x = -3$ The solution is $\enspace x = -3, \enspace y = 9$.